A Naive Question of Canonical Quantization A Real Scalar Field When we use canonical quantization method to quantize a free real scalar field in Schrodinger Picture.
The free real scalar Lagrangian is
$$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi^2$$
The EOM derived from this Lagrangian is 
\begin{align} 
(\partial_\mu\partial^{\mu} + m^2)\phi = 0 
\end{align}
If we do the Fourier transform from $\phi(\vec{x}, t)$ to $\phi(\vec{p}, t)$ as
\begin{align} 
\phi(\vec{p}, t) = \int \mathrm{d}^3\vec{x}  \phi(\vec{x}, t)e^{-i \vec{p}\cdot\vec{x}}\\
\phi(\vec{x}, t) = \int \frac{\mathrm{d}^3\vec{p}}{(2\pi)^3}  \phi(\vec{p}, t)e^{i \vec{p}\cdot\vec{x}}
\end{align}
Then the EOM in momentum space is written as
$$\left[ \frac{\partial^2}{\partial t^2} + (\vec{p}^2 + m^2) \right]\phi(\vec{p}, t) = 0 $$
Thus, for each value of $\vec{p}$, $\phi(\vec{p}, t)$solves this equation simple harmonic oscillator vibrating at frequency:
$$ \omega_{\vec{p}} = \sqrt{\vec{p}^2 + m^2} $$
When I quantize a Simple Harmonic Oscillator in Quantum Mechanics as defining
$$ a = \sqrt{\frac{\omega}{2}}q + \sqrt{\frac{i}{2\omega}}p \quad ~, \quad a^{\dagger} = \sqrt{\frac{\omega}{2}}q - \sqrt{\frac{i}{2\omega}}p $$
And
$$
q = \sqrt{\frac{1}{2\omega}}(a + a^{\dagger}) \quad ~, \quad p = - i \sqrt{\frac{\omega}{2}}(a - a^{\dagger})
$$
And if we do the similiar thing in QFT as writing
$$
\phi(\vec{p}) = \sqrt{\frac{1}{2\omega_{\vec{p}}}}(a_{\vec{p}} + a^{\dagger}_{\vec{p}})
$$
It's manifestly wrong( Because we need a real $\phi(\vec{x})$, which needs $\phi^{*}(\vec{p}) = \phi(-\vec{p})$). I want to know why we need to write the $\phi(\vec{p})$ as
$$
\phi(\vec{p}) = \sqrt{\frac{1}{2\omega_{\vec{p}}}}(a_{\vec{p}} + a^{\dagger}_{-\vec{p}})
$$
By the way, I know the 
$$
\int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{\sqrt{2 \omega_{\vec{p}}}}\left[a_{\vec{p}}+a_{-\vec{p}}^{\dagger} \right] e^{i \vec{p} \cdot \vec{x}} = \int \frac{d^{3} p}{(2 \pi)^{3}} \frac{1}{\sqrt{2 \omega_{\vec{p}}}}\left[a_{\vec{p}} e^{i \vec{p} \cdot \vec{x}}+a_{\vec{p}}^{\dagger} e^{-i \vec{p} \cdot \vec{x}}\right]
$$
 A: If the field $\phi(x)$ is assumed to be real as it is in this question a property of Fourier transforms (FT) requires, that the FT of a real function complies with the symmetry property:
$$ \tilde{\phi^\ast}(p) = \tilde{\phi}(-p)$$
This can be checked be consulting properties of Fourier transforms. For complex functions it is no longer valid.
EDIT:
If the field operator is now as chosen as:
$$\tilde{\phi}(\vec{p}) = \sqrt{\frac{1}{2\omega_\vec{p}}}( a_\vec{p} + a^\dagger_{-\vec{p}})$$ 
it can be easily seen that $\tilde{\phi}^{\dagger}(\vec{p}) =\tilde{\phi}(-\vec{p})$  (the above identity for fourier coefficients translates in the formalism of operators to a replacement of $\ast \rightarrow \dagger$) is fulfilled as required:
$$\tilde{\phi}^\dagger(\vec{p}) = \sqrt{\frac{1}{2\omega_\vec{p}}}( a^\dagger_\vec{p} + a_{-\vec{p}})$$
and 
$$\tilde{\phi}(-\vec{p}) = \sqrt{\frac{1}{2\omega_\vec{p}}}( a_{-\vec{p}} + a^\dagger_\vec{p})$$
$\omega_\vec{p}$ does not change as $\vec{p}$ enters in its formula quadratically.
